\(\int e^{-\log ^2(25 x)} (e^x (-2 x+29 x^2+10 x^3)+e^x (2 x-20 x^2) \log (25 x)) \, dx\) [8116]

Optimal result

Integrand size = 47, antiderivative size = 21 \[ \int e^{-\log ^2(25 x)} \left (e^x \left (-2 x+29 x^2+10 x^3\right )+e^x \left (2 x-20 x^2\right ) \log (25 x)\right ) \, dx=e^{x-\log ^2(25 x)} x^2 (-1+10 x) \]

[Out]

x^2*(10*x-1)/exp(ln(25*x)^2)*exp(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {2326} \[ \int e^{-\log ^2(25 x)} \left (e^x \left (-2 x+29 x^2+10 x^3\right )+e^x \left (2 x-20 x^2\right ) \log (25 x)\right ) \, dx=x \left (x-10 x^2\right ) \left (-e^{x-\log ^2(25 x)}\right ) \]

[In]

Int[(E^x*(-2*x + 29*x^2 + 10*x^3) + E^x*(2*x - 20*x^2)*Log[25*x])/E^Log[25*x]^2,x]

[Out]

-(E^(x - Log[25*x]^2)*x*(x - 10*x^2))

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = -e^{x-\log ^2(25 x)} x \left (x-10 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int e^{-\log ^2(25 x)} \left (e^x \left (-2 x+29 x^2+10 x^3\right )+e^x \left (2 x-20 x^2\right ) \log (25 x)\right ) \, dx=e^{x-\log ^2(25 x)} x^2 (-1+10 x) \]

[In]

Integrate[(E^x*(-2*x + 29*x^2 + 10*x^3) + E^x*(2*x - 20*x^2)*Log[25*x])/E^Log[25*x]^2,x]

[Out]

E^(x - Log[25*x]^2)*x^2*(-1 + 10*x)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
risch	\(\left (10 x -1\right ) x^{2} {\mathrm e}^{-\ln \left (25 x \right )^{2}+x}\)	\(21\)
parallelrisch	\(\left (10 \,{\mathrm e}^{x} x^{3}-{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-\ln \left (25 x \right )^{2}}\)	\(26\)

[In]

int(((-20*x^2+2*x)*exp(x)*ln(25*x)+(10*x^3+29*x^2-2*x)*exp(x))/exp(ln(25*x)^2),x,method=_RETURNVERBOSE)

[Out]

(10*x-1)*x^2*exp(-ln(25*x)^2+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int e^{-\log ^2(25 x)} \left (e^x \left (-2 x+29 x^2+10 x^3\right )+e^x \left (2 x-20 x^2\right ) \log (25 x)\right ) \, dx={\left (10 \, x^{3} - x^{2}\right )} e^{\left (-\log \left (25 \, x\right )^{2} + x\right )} \]

[In]

integrate(((-20*x^2+2*x)*exp(x)*log(25*x)+(10*x^3+29*x^2-2*x)*exp(x))/exp(log(25*x)^2),x, algorithm="fricas")

[Out]

(10*x^3 - x^2)*e^(-log(25*x)^2 + x)

Sympy [A] (verification not implemented)

Time = 7.90 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int e^{-\log ^2(25 x)} \left (e^x \left (-2 x+29 x^2+10 x^3\right )+e^x \left (2 x-20 x^2\right ) \log (25 x)\right ) \, dx=\left (10 x^{3} e^{- \log {\left (25 x \right )}^{2}} - x^{2} e^{- \log {\left (25 x \right )}^{2}}\right ) e^{x} \]

[In]

integrate(((-20*x**2+2*x)*exp(x)*ln(25*x)+(10*x**3+29*x**2-2*x)*exp(x))/exp(ln(25*x)**2),x)

[Out]

(10*x**3*exp(-log(25*x)**2) - x**2*exp(-log(25*x)**2))*exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int e^{-\log ^2(25 x)} \left (e^x \left (-2 x+29 x^2+10 x^3\right )+e^x \left (2 x-20 x^2\right ) \log (25 x)\right ) \, dx={\left (10 \, x^{3} - x^{2}\right )} e^{\left (-4 \, \log \left (5\right )^{2} - 4 \, \log \left (5\right ) \log \left (x\right ) - \log \left (x\right )^{2} + x\right )} \]

[In]

integrate(((-20*x^2+2*x)*exp(x)*log(25*x)+(10*x^3+29*x^2-2*x)*exp(x))/exp(log(25*x)^2),x, algorithm="maxima")

[Out]

(10*x^3 - x^2)*e^(-4*log(5)^2 - 4*log(5)*log(x) - log(x)^2 + x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int e^{-\log ^2(25 x)} \left (e^x \left (-2 x+29 x^2+10 x^3\right )+e^x \left (2 x-20 x^2\right ) \log (25 x)\right ) \, dx=10 \, x^{3} e^{\left (-\log \left (25 \, x\right )^{2} + x\right )} - x^{2} e^{\left (-\log \left (25 \, x\right )^{2} + x\right )} \]

[In]

integrate(((-20*x^2+2*x)*exp(x)*log(25*x)+(10*x^3+29*x^2-2*x)*exp(x))/exp(log(25*x)^2),x, algorithm="giac")

[Out]

10*x^3*e^(-log(25*x)^2 + x) - x^2*e^(-log(25*x)^2 + x)

Mupad [B] (verification not implemented)

Time = 12.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81 \[ \int e^{-\log ^2(25 x)} \left (e^x \left (-2 x+29 x^2+10 x^3\right )+e^x \left (2 x-20 x^2\right ) \log (25 x)\right ) \, dx=-\frac {{\mathrm {e}}^{-{\ln \left (x\right )}^2-4\,{\ln \left (5\right )}^2}\,\left (x^2\,{\mathrm {e}}^x-10\,x^3\,{\mathrm {e}}^x\right )}{x^{4\,\ln \left (5\right )}} \]

[In]

int(exp(-log(25*x)^2)*(exp(x)*(29*x^2 - 2*x + 10*x^3) + log(25*x)*exp(x)*(2*x - 20*x^2)),x)

[Out]

-(exp(- log(x)^2 - 4*log(5)^2)*(x^2*exp(x) - 10*x^3*exp(x)))/x^(4*log(5))